Can't wrap my head around these (special) product/quotient rules.

[rhombuster]

I've been given a couple questions that just aren't making sense to me.

In general, the derivative of a product is not the product of the derivatives. Find nonconstant functions f and g such that the derivative of fg equals f'g'.

I have the same question but for the quotient rule. (f/g equals f'/g')

I can't find any information on this.

I'm pretty sure that I'd have to use e^x for the product rule and ln(something) for the quotient. I can't think of any functions that would end up working like this.

I know the derivative of e^x = e^x. Can anyone help me with this. Everywhere on the net just posts proofs for this being unable to happen.

 

[soroban]

Hello, rhombuster!

In general, the derivative of a product is not the product of the derivatives. Find nonconstant functions \(\displaystyle f\) and \(\displaystyle g\) such that the derivative of \(\displaystyle fg\) equals \(\displaystyle f'g'.\)I have the same question but for the quotient rule: \(\displaystyle (\frac{f}{g})' \:=\:\frac{f'}{g'}\)

I played with this concept while in college.This is what I found . We want: \(\displaystyle (f\cdot g)' \:=\:f'\cdot g'\)That is: \(\displaystyle fg' + f'g \:=\:f'g'\)Solve for \(\displaystyle g'\!:\;f'g' - fg' \:=\:f'g \quad\Rightarrow\quad (f'-f)g' \:=\:f'g \quad\Rightarrow\quad \dfrac{g'}{g} \:=\:\dfrac{f'}{f'-f} \)Integrate: \(\displaystyle \displaystyle\;\int\frac{g'}{g} \:=\:\int\frac{f'}{f'-f} \quad\Rightarrow\quad \ln|g| \:=\: \int\frac{f'}{f'-f}\) Select a function for \(\displaystyle f(x)\) and determine \(\displaystyle g(x).\)Example: \(\displaystyle f \,=\,x^2 \quad\Rightarrow\quad f' \,=\,2x\)Then: \(\displaystyle \dfrac{f'}{f'-f} \:=\:\dfrac{2x}{2x-x^2} \:=\: \dfrac{2}{2-x}\;\text{ for }x \ne 0,2\)We have: \(\displaystyle \displaystyle\;\ln|g| \;=\;\int\frac{2\,dx}{2-x} =\;-2\ln(2-x) + C \;=\;\ln\left[C(2-x)^{-2}\right] \)Hence: \(\displaystyle g \;=\;C(2-x)^{-2}\)The two functions are: .\(\displaystyle f \,=\,x^2\,\text{ and }\,g \,=\,\dfrac{C}{(2-x)^2}\)

 

[rhombuster]

We haven't been taught integration and I still can't figure this question out. Either way, I appreciate the time you guys took to answer this, have a good one.